TAS FET EUCLIDEAN GEOMETRY COURSE
BOOKLET 3
Circle Geometry
S
A
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by The TAS Maths Team
Circle Geometry
Geometry FET
Course Booklets Set
WWW.THEANSWER.CO. ZA
"SUBTEND" . . .
CIRCLE GEOMETRY
P
P
O
O
Understand the word!
The Language (Vocabulary)
GROUP 1 AND
Centre
radius
diameter
centre
A
A
A
B
Figure 2
S
A
T
O
A
180º
Figure 3
ˆ
a central AOB
at the centre of the circle, and
ˆ
an inscribed APB
at the circumference of the circle.
Chords
B
To ensure that you grasp the meaning of the word 'subtend' :
B
Arcs (major & minor)
A
sector
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P
O
move along the radii to meet at O and back ; then,
minor arc AB
move to meet at P on the circumference and back.
Turn your book upside down and sideways.
You need to recognise different views of these situations.
Segments (major & minor)
major
segment
A
A
Consider that
subtend
means support.
Take each of the figures:
chord AB
Place your index fingers on A & B ;
O
Figure 4
major arc AB
A
Sectors
O
B
or chord AB, subtends:
In all the figures, arc AB (AB),
Circumference
chord
B
Central and Inscribed angles
B
O
B
Figure 1
C
Diameter
Radius
2
A
P
P
minor
segment
Take note of whether the angles are acute, obtuse, right, straight or reflex.
Redraw figures 1 to 4 leaving out the chord AB completely and observe
the arc subtending the central and inscribed angles in each case.
B
B
1
GROUP
GROUP
3
Cyclic Quadrilaterals
Tangents
A cyclic quadrilateral is a quadrilateral which has all 4 vertices on the circumference
of a circle.
Special lines
D
4
A tangent is a line which touches a circle at a point.
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Points A, B, C and D
are concyclic,
i.e. they lie on the same circle.
O
A
tangent
point of contact
C
B
Note: Quadrilateral AOCB is not a cyclic quadrilateral because point O
is not on the circumference ! (A, O, C and B are not concyclic)
A secant is a line which cuts a circle (in two points).
We name quadrilaterals by going around, either way, using
consecutive vertices, i.e. ABCD or ADCB, not ADBC.
secant
Exterior angles of polygons
The exterior angle of any polygon is an angle which is formed between one side of
the polygon and another side produced.
e.g. A triangle
e.g. A quadrilateral /
cyclic quadrilateral
A
E
D
B
A
C
D
B
NB : It is assumed that the
tangent is perpendicular
C
ˆ is an exterior ø of ΔABC.
ACD
[NB : BCD is a straight line ! ]
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to the radius (or diameter)
ˆ is an exterior ø of c.q. ABCD.
ADE
at the point of contact.
[NB : CDE is a straight line ! ]
2
SUMMARY OF CIRCLE GEOMETRY THEOREMS
I
II
III
x
The
'Centre'
group
The
' No Centre '
group
The
' Cyclic Quad.'
group
2x
2x
x
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x
Equal
chords !
x
There are ' 3 ways to
prove that a quad. is
a cyclic quad '.
y
x + y = 180º
IV
The
'Tangent'
group
Equal
tangents !
There are ' 2 ways to prove that
a line is a tangent to a ?'.
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3
Equal
radii !
GROUPING OF CIRCLE GEOMETRY THEOREMS
The grey arrows indicate how various theorems are used to prove subsequent ones
I
II
III
x
The
'Centre'
group
The
' No Centre '
group
The
' Cyclic Quad.'
group
2x
2x
x
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Equal
chords !
x
Equal chords subtend equal angles
and, vice versa,
equal angles are subtended
by equal chords.
x
There are ' 3 ways to
prove that a quad. is
a cyclic quad '.
y
x + y = 180º
IV
The
'Tangent'
group
Equal
tangents !
There are ' 2 ways to prove that
a line is a tangent to a ?'.
Copyright © The Answer Series
Equal
radii !
4
2
GEOMETRY
The angle which an arc of a circle subtends at the centre is
double the angle it subtends at any point on the circumference.
Given :
ˆ at the centre and APB
ˆ at the circumference.
O, arc AB subtending AOB
P
ˆ = 2APB
ˆ
AOB
To prove :
Construction : Join PO and produce it to Q.
Proof :
Proofs required for
Pˆ 1 =
Let
x
Then  = x . . . s opp = radii
 Oˆ 1 = 2x . . . exterior  of DAOP
Similarly, if Pˆ 2 = y, then Oˆ 2 = 2y
exam purposes
12
x y
O
1 2
x
y
A
Q
ˆ = 2x + 2y = 2(x + y) = 2 APB
ˆ
 AOB
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A
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3
B
The opposite angles of a cyclic quadrilateral are supplementary.
A
x
Given : O and cyclic quadrilateral ABCD
To prove :
 Circle Geometry Theorems
1
Proof :
Construction : Join OA and OB
In Ds OPA & OPB
. . . given
(3) OP is common
 DOAP º DOBP
. . . RHS
 AP = PB, i.e. OP bisects chord AB
A
1 2
P
D
1
B
. . .  at centre = 2 
 at circumference
= 360º – 2x . . . s about point O
C
4
B
quadrilateral = 360º
The angle between a tangent to a circle and a chord drawn from the point of
contact is equal to the angle subtended by the chord in the alternate segment.
ˆ ,
Given : Tangent AC and chord AB subtending APB
with C & P on opposite sides of AB.
ˆ = APB
ˆ ,
A
1
To prove :
Construction : Draw diameter AD and join DP.
Proof : We have
ˆ
= 90º
DAC
. . . diameter ⊥ tangent
&
ˆ
= 90º
DPA
. . .  in semi-
∴ Aˆ 1 + Aˆ 2 = 90º
See Gr 11 Maths 3 in 1
Study Guide
and
But
ˆ
ˆ
A
2 = P1
5
D
P
B
1
2
2
1
x
A
Pˆ 1 + Pˆ 2 = 90º
. . . s in same seg.
ˆ
∴ Aˆ 1 = Pˆ 2 ( = APB)
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O
x
 Ĉ = 1 (360º – 2x) = 180º – x . . .  at centre = 2 
2
 at circumference
 Â + Ĉ = 180º
s
sum of the  of a
...
&  B̂ + D̂ = 180º
O
. . . radii
2
ˆ = 2x
O
1
 Oˆ 2
To prove : AP = PB
(2) Pˆ 1 = Pˆ 2 (= 90º)
 =
Let
Then
Given : O with OP ⊥ AB
(1) OA = OB
B̂ + D̂ = 180º
Construction : Join BO and DO.
The line segment drawn from the centre of a circle,
perpendicular to a chord, bisects the chord.
Proof :
 + Ĉ = 180º &
Remember :
Always give
REASONS for
each statement !
C
PROVING THEOREMS
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x y
y
x
2x
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6
x y
2y
x
y
2x 2y
2x
2y
y
x
? THEOREM PROOFS: A Visual presentation
The Situation
Examinable
Theorem
Statements
The LOGIC . . .
Construction
radii
1
centre
chord
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Examin
2
radius produced
inscribed ∠
central ∠
y
y
x
arc
x 2x 2y
equal base ∠ s
radii
3
xy
x y
centre
exterior ∠ s of Δs
y
Circle Theorem Proofs
The line drawn
from the centre
of a circle,
perpendicular
to a chord,
bisects the chord.
x
x
opposite ∠ s in a
cyclic quadrilateral
2y
y
2x
∠ at centre = 2 % ∠ at circumference
& ∠ s about a point = 360º.
Note: The bolded words in the statements are the approved 'reasons' to use.
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7
The angle which an
arc of a circle subtends
at the centre is
double the angle
it subtends at
any point on the
circumference.
The opposite angles
of a cyclic
quadrilateral
are supplementary.
? THEOREM PROOFS: A Visual presentation, continued
The Situation
Method 1:
4
inscribed ∠
diameter & join . . .
chord
x
Method 2:
radii
x
Theorem Statement
The angle between a tangent to a circle
and a chord drawn from the point of contact
is equal to the angle subtended by the chord
in the alternate segment.
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90° –
x
x
90° –
x
x
x
∠ in the
same segment
90° – x
x
x
Tan Chord theorem
x
2x
2x
x
x
s
∠ at centre = 2 % ∠ at circumference
sum of ∠ of Δ
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8
90° – x
base ∠ s of isosceles Δ
radius ⊥ tangent
This theorem is known as the
x
s
diameter ⊥ tangent
& ∠ in semi-?
tangent
Circle Theorem Proofs
The LOGIC . . .
Construction
Examinable
FURTHER ? THEOREM PROOFS: A Visual presentation
The Situation
5
?
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180º
diameter
Central angle is
a straight angle
6
The LOGIC . . .
x
x
(at the
circumference
of the circle)
arc
7
2x
2x
∠ subtended by an arc (or chord)
(at the centre of the circle)
∠ s subtended by an arc (or chord)
(at the circumference)
x
cyclic
quadrilateral
x
∠ s subtended by
the same arc
are equal.
The exterior ∠ of a
cyclic quadrilateral
exterior
angle
y
y
Adjacent ∠ s :
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The ∠ in a semi-?
is a right ∠.
Inscribed angle is
a right angle
Construction: radii
∠ s subtended
by an arc
Circle Theorem Proofs
Theorem
Statements
The LOGIC . . .
9
x
x + y = 180º
= the interior opposite ∠.
FURTHER ? THEOREM PROOFS: A Visual presentation, continued . . .
The Situation
The LOGIC . . .
Construction
Method 1:
radii
8
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radii ⊥ tangents
Method 2:
Theorem
Statement
congruent Δs
chord and inscribed ∠
x
x
x
x
Circle Theorem Proofs
x
x
tan-chord theorem 4
Proofs 5
to 8
sides opposite
equal base ∠ s in Δ
are not examinable, but,
the LOGIC is crucial when studying geometry.
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x
10
Tangents to a circle
from a common point
are equal.
Again, the
bolded words
are the ‘approved
reasons' to use.
EXAMPLE 1
PROBLEM-SOLVING: An Active Approach
This is an excellent example testing so many
?-geometry theorems (excluding tangents).
ACT !
A
Be Active . . .
Mark all the information on the drawing :
equal or parallel sides equal øs right øs
K
Use all the Clues . . .
intersecting lines adj. suppl. & vert. opp. øs ; øs about a point.
ˆ
(c) M
4
ˆ +N
ˆ
(d) N
1
2
parallel lines alternate, corresponding & co-interior øs
ˆ
(e) M
1
(f) Prove that KG = GM
equal chords equal øs
90º angle the Theorem of Pythagoras, or . . .
sum of the interior øs of a quadrilateral = 360º ;
types & properties of all quadrilaterals
F
76º
K
1
O
2
1
3 2
H
G
4
ˆ
ˆ = 1O
(c) M
4
1
12
N
2
ˆ +N
ˆ = 180º - 76º
(d) N
1
2
= 104º . . . opp. øs of c.q. are suppl.
ˆ = 90º
(e) KML
. . . ø in semi-?
ˆ = 38º in 4.3
ˆ &M
ˆ
â M 1 = 180º - (90º + 38º) . . . str. NMP
4
= 52º Previous geometry
Revise previous geometry under the headings
angles lines triangles quadrilaterals
(f)
ˆ = corresponding KML
ˆ = 90º
G
3
. . . ON || LM
i.e. line from centre, OG ⊥ chord KM
Circle geometry
â KG = GM Recall and apply the facts by group.
11
L
1 2
3 2 1
= 38º . . . ø at centre = 2 % ø at circumference
Recall the Theory systematically . . .
Copyright © The Answer Series
M
N
ˆ = corr. Lˆ . . . ON || LM
(b) O
1
1
= 76º isosceles Δs ; congruency, similarity, theorem of Pythagoras
s
1 2
(a) Lˆ 1 ( = Fˆ )
= 76º . . . øS in same segment
Δs sum of the interior øs = 180º ; exterior ø of Δ ; equilateral Δs ;
3 2 1
Answers
There are
3 theorems
in ? geometry
which involve
right angles.
diameter ø in semi-? = 90º ; diameter ⊥ tangent
P
1 2
4
ˆ
(b) O
1
L
H
1
3 2
G
(a) Lˆ 1
equal radii or tangents (in triangles) equal base øs
O
2
1
S
A
T
Calculate, giving reasons,
the sizes of :
The information provides clues to facts :
T
76º
ON || LM and F̂ = 76º.
radii diameter tangents
C
F
O is the centre of the circle
and diameter KL is produced
to meet NM produced at P.
M
P
EXAMPLE 2
3 ways to prove that a quadrilateral is cyclic
(a) Complete the following by writing the appropriate missing word.
If a chord of a circle subtends a right angle on the circumference,
then this chord is a . . . . . .
B
(b) A, B, C and D are points on the circle.
1
1
1
If M̂ = N̂, prove that AC is a diameter of the circle.
Answers
z1
y2
. . . given
ˆ = C
ˆ = y . . . vert. opp. øs
& C
1
2
Prove that a pair of
opposite angles
is supplementary.
N
1
x+y
Then :
Bˆ 1 = x + y
. . . ext. ø of ΔBMC
ˆ = x+y
& D
1
. . . ext. ø of ΔDCN
ˆ = 180º
But, Bˆ 1 + D
1
â 2(x + y) = 180º
â x + y = 90º
ˆ
ˆ = 90º
i.e. B = D
x
B
A
 + Ĉ = 180º
or B̂ + D̂ = 180º
D
1
. . . opp. ø of c.q.
7
Either prove :
1̂ = 2̂
4
8 3
6
1
y
2
C
or 3̂ = 4̂
or 5̂ = 6̂
or 7̂ = 8̂
2 ways to prove that a line is a tangent to a ?
2
x
N
We use the 2 converse theorem statements to prove that a line is a tangent to a ? if . . .
converse theorem
1
â AC is a diameter . . . subtends a right ø
ACT!
. . . it is perpendicular to a radius
at a point where the radius meets
the circumference.
O
Prove that OP ⊥ AB;
then AB will be a tangent
A
P
B
converse theorem
A: Be Active
. . . when drawn through
the end point of a chord,
it makes with the chord
an angle equal to an angle
A
in the alternate segment.
C: Use all your Clues
T: Apply the Theory systematically, recalling
each group, and fact, one at a time.
Copyright © The Answer Series
5
2
C
D
Either prove :
A
Prove that an exterior angle
of a quadrilateral is equal
to the interior opposite angle.
M
Either prove :
x1 = x2 or y1 = y2
or z1 = z2 or p1 = p2
y1
x+y
s
2
z2
p2
B
converse theorem
(a) . . . diameter.
p1 x
2
y1
converse theorem
2
S
A
T
D
x1
Prove that one side subtends
equal angles at the two other points
(on the same side).
C
A
AB produced and DC produced
meet at M.
1
converse theorem
2
2
BC produced and AD produced
meet at N.
(b) Let M̂ = N̂ = x
We use the 3 converse statements for cyclic quadrilaterals.
M
12
y
x
B
Prove that
x = y;
then AB
will be
a tangent
Note :
It is a good idea to
draw a faint circle
(as shown) to see
this converse
theorem clearly.
EXAMPLE 3 (National November 2017 P2, Q10)
(b)
ˆ
= 90º . . . tangent ⊥ radius
OWS
ˆ
ˆ
â OVN = OWS
(= 90º)
(i)
In the diagram, W is a point on the
circle with centre O.
â MN || TS
. . . corresp. øs equal
V is a point on OW.
(ii)
Chord MN is drawn such that
MV = VN.
The tangent at W meets
OM produced at T and
ON produced at S.
O
T
M
1
2 1
(a) Give a reason why OV ⊥ MN.
(b) Prove that:
(i)
MN || TS
(ii)
TMNS is a cyclic quadrilateral
V
W
2
O
S
A
T
M
T
1
1 2
1
2
Shade the
quadrilateral
TMNS
V
2 N
W
S
(iii) OS . MN = 2ON . WS (a Grade 12 question)
Answers
The most 'basic'
way to prove
lines || is:
alt. or corresp.
øs equal or
co-int. øs suppl.
1
2 N
S
There are 3 ways to prove that a quadrilateral is
a cyclic quadrilateral – choose 1:
(a) Line (OV) from centre to midpoint of chord (MN) y
x
In this case, the midpoint of the chord is given, and
we can conclude that OV ⊥ MN because of that.
Prove that:
x + y = 180º
Prove that:
Ext. ø = int. opp. ø
s
ˆ = N
ˆ
M
1
1 . . . ø opposite equal radii
Note: Analyse the information and the diagram.
= Ŝ
So far, we have used and applied a 'centre' theorem, in (a).
Another clue is the 'tangent' at W.
Think about tangent facts . . . .
. . . corresp. øs ; MN || TS
â TMNS is a cyclic quadrilateral
Copyright © The Answer Series
13
Prove that:
A side subtends equal øs
at 2 other vertices
We chose and proved
that the exterior ø of
quadrilateral TMNS
= the interior opposite ø
converse ext. ø
. . . of cyclic quad.
(iii)
EXAMPLE 4
This question looks like ratio and proportion.
Mark the sides on the diagram.
Don't be put off by this drawing !
Direct your focus to one situation at a time ☺
The sides appear to involve ΔOWS, which has VN || WS,
(even though MN = 2VN)
. . . Maybe apply the proportion theorem in this Δ?
A
But, the sides in the question involve the horizontal sides
WS and VN.
1 2
So, proportion theorem is excluded.
We will use similar Δs !
S
A
T
X
In Δs OVN and OWS
ˆ is common
O
2
ˆ = OWS
ˆ
OVN
1
2
1
â ΔOVN ||| ΔOWS
. . . øøø
B
2
2
Q
Y
3
P
. . . corresp. øs ; MN || TS
1
1 2
T
Make statements, with reasons,
Let's 'arrange' the sides to suit the question.
OS
â
= WS ⎛⎜ = OW ⎞⎟
ON
VN ⎝ OV ⎠
. . . equiangular Δ
1. In ?XPBA : about P̂1 and B̂1
s
ˆ
2. In ?ABYQ : about B̂1 and AQY
â OS . VN = ON . WS
But VN =
1
MN
2
ˆ
3. In quadrilateral APTQ : about P̂1 and AQT
. . . V midpoint MN
â OS . 1 MN = ON . WS
4. What can you conclude about quadrilateral APTQ?
2
Mark the øs on the drawing as you proceed.
% 2)â OS . MN = 2ON . WS Answers
O
M
T
2
1 2
1
V
1
W
2
. . . arc XA subtends øs in same segment
ˆ
2. In c.q. ABYQ : B̂1 = AQY
. . . exterior ø of cyclic quad.
ˆ
3. In quad. APTQ : P̂1 = AQT
N
4. ∴ APTQ is a cyclic quad.
S
Copyright © The Answer Series
1. In c.q. XPBA : P̂1 = B̂1
14
. . . both = Bˆ1 above
. . . converse of exterior ø of cyclic quad.
EXAMPLE 5
EXAMPLE 6
P
Prove that PA
is a tangent to M.
8
AB is a tangent to the circle at B and BD is a chord.
T
AD cuts the circle in E.
12
C is a point on BD so that ABCE is a cyclic quadrilateral.
M
3
A
4
AC, BE and CE are joined.
B
Q
A
1 2
Answers
In right-d ∆MAQ :
AM = 5 units
. . . 3 :4 :5 ∆ ; Pythag.
. . . TM = AM = radii
 TM = 5 units
 PM = 13 units
P
8
12
S
A
T
B
A
i.e. PM 2 = PA2 + AM 2
1
2
2
M
3
 ∆PAM is a 5 : 12 : 13 ∆ !
ˆ = 90º
 PAM
T
1
4
Q
3
E
3 21
C
D
B
Prove that :
(a) AB = AC
. . . converse of Pythag.
(b) AC is a tangent to circle ECD at point C.
 PA is a tangent to  M  . . . converse tan chord theorem
Answers
B̂1 = x
(a) Let
then D̂ = x
. . . tan chord theorem
and Ĉ2 = x
. . . s in same segment
&
See Gr 11 Maths 3 in 1
Study Guide
A
1 2
y
Let B̂2 = y
then
 2 = y
Ĉ3 = D̂ + Â 2
. . . s in same segment
B
∴ AB = AC
Module 9b: Circle Geometry
 Notes
 Exercises
 Full Solutions
(b) Ĉ2 = D̂
. . . B̂1 = x and
. . . converse tan chord thm
∴ AC is a tangent to ECD at C.
Copyright  The Answer Series
15
B̂2 = y
. . . sides opp = s
. . . both = x in (a)
1
x
2
y
x
3 21
C
. . . ext.  of D
= x+y
ˆ = x+y
& ABC
1
2
3
E
x
D
ACCEPTABLE REASONS: CIRCLE GEOMETRY
THEOREM STATEMENT
O
O
O
O
O
x
O
2x
ACCEPTABLE REASON(S)
The tangent to a circle is perpendicular
to the radius/diameter of the circle at
the point of contact.
tan ⊥ radius
tan ⊥ diameter
If a line is drawn perpendicular to a
radius/diameter at the point where the
radius/diameter meets the circle, then
the line is a tangent to the circle.
line ⊥ radius OR
The line drawn from the centre of a
circle to the midpoint of a chord is
perpendicular to the chord.
The line drawn from the centre of a
circle perpendicular to a chord bisects
the chord.
The perpendicular bisector of a
chord passes through the centre of
the circle.
The angle subtended by an arc at the
centre of a circle is double the size of
the angle subtended by the same arc
at the circle (on the same side of the
chord as the centre)
O
The angle subtended by the diameter
at the circumference of the circle
is 90º.
O
If the angle subtended by a chord at
the circumference of the circle is 90°,
then the chord is a diameter.
Copyright © The Answer Series
THEOREM STATEMENT
x
x
y
y
x
Angles subtended by a chord of the
circle, on the same side of the
chord, are equal
øs in the same seg
If a line segment joining two points
subtends equal angles at two
points on the same side of the line
segment, then the four points are
concyclic.
line subtends equal øs
S
A
T
converse tan ⊥ radius OR
x
converse tan ⊥ diameter
ACCEPTABLE REASON(S)
OR
converse øs in the same seg
(This can be used to prove that the
four points are concyclic).
line from centre to midpt of chord
x
x
Equal chords subtend equal angles
at the circumference of the circle.
equal chords; equal øs
Equal chords subtend equal angles
at the centre of the circle.
equal chords; equal øs
line from centre ⊥ to chord
x
xO
perp bisector of chord
x
øat centre
= 2 % ø at circumference
x
A
x
B
x
Equal chords in equal circles subtend
equal angles at the circumference of
the circles.
Equal chords in equal circles
subtend equal angles at the centre
of the circles.
(A and B indicate the centres of
the circles)
øs in semi circle OR
diameter subtends right angle
OR ø in ½ ?
x
chord subtends 90º OR
y
The opposite angles of a cyclic
quadrilateral are supplementary
(i.e. x and y are supplementary)
180º – x
If the opposite angles of a
quadrilateral are supplementary
then the quadrilateral is cyclic.
x
converse øs in semi circle
16
equal circles; equal chords;
equal øs
equal circles; equal chords;
equal øs
opp øs of cyclic quad
opp øs quad sup OR
converse opp øs of cyclic quad
THEOREM STATEMENT
The exterior angle of a cyclic
quadrilateral is equal to the
interior opposite angle.
x
x
If the exterior angle of a quadrilateral
is equal to the interior opposite
angle of the quadrilateral, then the
quadrilateral is cyclic.
x
x
B
Two tangents drawn to a circle
from the same point outside the
circle are equal in length (AB = AC)
A
C
x
x
y
x
y
a
b
y
ACCEPTABLE REASON(S)
ext ø of cyclic quad
ext ø = int opp ø
OR
converse ext ø of cyclic quad
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Tans from common pt
OR
Tans from same pt
The angle between the tangent to
a circle and the chord drawn from
the point of contact is equal to the
angle in the alternate segment.
tan chord theorem
If a line is drawn through the endpoint of a chord, making with the
chord an angle equal to an angle
in the alternate segment, then the
line is a tangent to the circle.
converse tan chord theorem
(If x = b or if y = a then the
line is a tangent to the circle)
Copyright © The Answer Series
NOTES
OR
ø between line and chord
17
PLEASE NOTE
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These Geometry materials (Booklets 1 to 4) were created and
produced by The Answer Series Educational Publishers (Pty) (Ltd)
to support the teaching and learning of Geometry in high schools
in South Africa.
They are freely available to anyone who wishes to use them.
This material may not be sold (via any channel) or used for
profit-making of any kind.
GRADES 8 - 12
ALL MAJOR SUBJECTS IN
ENGLISH & AFRIKAANS
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